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I'm currently a student looking through the differentiation topic and came across this problem as follows...

$$V=125\pi r-\pi r^3$$

Where $ r$ is a variable and $r^3$ represents r cubed.

The first derivative $\frac {dV}{dr} = 125\pi-3r^2 \pi$

The second derivative $\frac {d^2V}{dr^2} = -6r\pi$

The formula and derivatives are both correct, however, my original attempt at them was the exact same apart from the exclusion of $\pi$.

If $y=57$ were to be differentiated the result would be $0$ according to Wallis' law. But as$\pi$ is simply a constant real number, why does it not simply disappear? (For lack of a better word.)

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    As you claim, $\pi$ is like any constant. $\frac d{dx} Cx=C$ for any constant $C$. – 2017-02-26
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    $\pi$ is a factor, not a summand, therefore it does not "disppear", but stays as a factor. This is irrelevant for the exercise here, but if the factor would be negative, you would determine minimum / maximum wrong, if you would omit it. – 2017-02-26

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Let us check it out by looking at what the definition says. There is nothing mysterious.

By differentiating the function $V = 125\pi r - \pi r^{3}$ (where $\pi$ is a constant like $125$ so the value of $V$ is dependent on $r$ alone) at a point $r$ we mean taking the limit $$ \lim_{h \to 0} \frac{[125 \pi (r+h) - \pi (r+h)^{3}] - [125 \pi r - \pi r^{3}]}{h}. $$ Now you see that $\pi$, like $125$, does not contribute anything to the differentiation of this function.

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    When you wanna hang out you got to take her out, cocaine...or are you not *that* Eric Clapton? – 2017-02-26